Testing for absolute convergence? $\frac{(-1)^n}{5n+1}$

Question

I'm trying to test the summation $\sum^\infty_{n=0}\frac{(-1)^n}{5n+1}$ for absolute convergence.

By alternating series test, I can tell is is at least conditionally convergent.

However, when I used the ratio test, I got 1, which means it doesn't tell us anything.

A google search showed an answer on yahoo answers using the limit comparison test, using the harmonic series to compare it with, but they seemed to ignore the $(-1)^n$...

The answer in the book says it is conditionally convergent, but I can't work out how to show that it is not absolutely convergent.

Any ideas?

Answer 1

Taking the absolute value and using the asymptotic comparison we find $$\frac1{5n+1}\sim_\infty \frac15\frac1n$$ that the series isn't absolutely convergent since the harmonic series isn't convergent.

Answer 2

You have:

$\left|\frac{(-1)^n}{5n+1} \right|=\frac{1}{5n+1}>\frac{1}{5n+5}$,

therefore:

$\sum\limits_{n=0}^{\infty}\left|\frac{(-1)^n}{5n+1} \right|>\sum\limits_{n=0}^{\infty}\frac{1}{5n+5}=\frac{1}{5}\sum\limits_{n=1}^{\infty}\frac{1}{n}=\infty$,

So the series is not absolutely convergent.

Answer 3

The absolute series would be: $\sum_{n=0}^\infty\frac{1}{5n+1}\geq\sum_{n=0}^\infty\frac{1}{6n}$ (at least for sufficiantly large $n$ it is). Then:$$\sum_{n=0}^\infty\frac{1}{6n}=\frac16\sum_{n=0}^\infty\frac{1}{n}$$ Wich is obviously divirgent.

Answer 4

Taking the absolute value, we can apply the integral test:

$$\int_1^\infty \! \frac{1}{5x+1} dx = \lim_{N \to \infty}\frac{1}{5}\Big(\ln(5N+1) - \ln(6)\Big)$$

Certainly, the natural log is unbounded. Therefore, the integral does not converge, and we know the integral converges $\iff$ the series converges.

We conclude the series is not absolutely convergent. However, it is conditionally convergent by the alternating series test.